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Q.
Molar conductivity of $NH_4OH$ can be calculated by the equation,
Electrochemistry
Solution:
$\Lambda^{\circ}_{Ba\left(OH\right)_2}=\Lambda^{\circ}_{Ba^{2+}}+2\Lambda^{\circ}_{OH^{-}}$
$\Lambda^{\circ}_{BaCl_2}=\Lambda^{\circ}_{Ba^{2+}}+2\Lambda^{\circ}_{Cl^{-}}$
$\Lambda^{\circ}_{NH_4Cl}=\Lambda^{\circ}_{NH^{+}_{4}}+\Lambda^{\circ}_{Cl^{-}}$
After substituting the above in
$\Lambda^{\circ}_{NH_4OH}=\frac{\Lambda^{\circ}_{Ba\left(OH\right)_2}+2\Lambda^{\circ}_{NH_4Cl}-\Delta^{\circ}_{BaCl_2}}{2}$
we get, $\Lambda^{\circ}_{NH_4OH}=\Lambda^{\circ}_{NH^{+}_{4}}+\Lambda^{\circ}_{OH^{-}}$